Fie ABCDEF un hexagon regulat si \( M\in (AC) , N\in (CE) \) astfel incat \( \frac{AM}{AC}=\frac{CN}{CE}=r \). Pentru ce valoare a lui r punctele B,M,N sunt coliniare?
O.I.M. 1982
Coliniaritate
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Marius Mainea
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\( \frac{AM}{MC}=\frac{r}{1-r}\ \Longrightarrow \overline{\underline{\left\|{\vec{BM}=(1-r)\vec{BA}+r\vec{BC}\right\| \)
\( \frac{CN}{NE}=\frac{r}{1-r}\ \Longrightarrow \vec{BN}=(1-r)\vec{BC}+r\vec{BE} \)
Ramane sa exprimam vectorul \( \vec{BE} \) in functie de \( \vec{BA} \) si \( \vec{BC} \)
\( \vec{BE}=\vec{BF}+\vec{BC}=(\vec{BC}+\vec{CF})+\vec{BC}=2\vec{BC}+2\vec{BA} \)
\( \Longrightarrow \overline{\underline{\left\|{\vec{BN}=2r\vec{BA}+(1+r)\vec{BC}\right\| \)
\( \vec{BM} \) si \( \vec{BN} \) coliniari \( \Longleftrightarrow \frac{1-r}{2r}=\frac{r}{1+r}\ \Longleftrightarrow r=\frac{1}{\sqrt{3}} \)
\( \frac{CN}{NE}=\frac{r}{1-r}\ \Longrightarrow \vec{BN}=(1-r)\vec{BC}+r\vec{BE} \)
Ramane sa exprimam vectorul \( \vec{BE} \) in functie de \( \vec{BA} \) si \( \vec{BC} \)
\( \vec{BE}=\vec{BF}+\vec{BC}=(\vec{BC}+\vec{CF})+\vec{BC}=2\vec{BC}+2\vec{BA} \)
\( \Longrightarrow \overline{\underline{\left\|{\vec{BN}=2r\vec{BA}+(1+r)\vec{BC}\right\| \)
\( \vec{BM} \) si \( \vec{BN} \) coliniari \( \Longleftrightarrow \frac{1-r}{2r}=\frac{r}{1+r}\ \Longleftrightarrow r=\frac{1}{\sqrt{3}} \)